Let $G$ be a group of order $n$, $H$ a subgroup of $G$ of order $m$, $k=\dfrac{n}{m}$ and $S_k$ the symmetric group on $k$ symbols.
(a) Show that there is a non-trivial group homomorphism $\phi : G\to S_k$.
(b) Assuming $\dfrac{k!}{2}<n$, show that $G$ has a non-trivial proper normal subgroup.
My attempt $:$
(a) Let $X$ be the set of all cosets of $H$ in $G.$ Then $|X| = [G : H] = \dfrac {|G|} {|H|} = \dfrac {n} {m} = k.$ Consider the left action of $G$ on $X$ by left multiplication i.e. $g. (aH) := (ga) H,$ for all $g \in G$ and $aH \in X.$ Then for all $g \in G,$ the map $\varphi_g : X \longrightarrow X$ defined by $$\varphi_g(aH) = (ga) H,\ aH \in X$$ is a bijection. Now it is easy to verify that $\varphi_{g_1g_2} = \varphi_{g_1} \circ \varphi_{g_2},$ for all $g_1,g_2 \in G.$ Hence the action of $G$ on $X$ induces a homomorphism $\phi : G \longrightarrow S(X)$ defined by $$\phi (g) = \varphi_g,\ g \in G.$$ Since $|X| = k,$ it follows that $S(X) \simeq S_k.$ So in order to prove (a) we need only to show that $\phi$ is non-trivial. If $\phi$ were trivial then $\varphi_g = \text {id}_X,$ for all $g \in G.$ But then $\varphi_g (H) = H,$ for all $g \in G.$ This implies that $gH = H,$ for all $g \in G.$ But then for all $g \in G$ we have $g \in H,$ showing that $G = H.$ This can't be possible if $m \lt n.$ So we are through for $m \lt n.$ In fact for $m = n,$ we have $k = 1.$ So there is no question of getting a non-trivial homomorphism when $m = n.$
(b) For proving (b) I am thinkng about the kernel of $\phi.$ Can it proved that $\phi$ fails to be injective if $\dfrac {k!} {2} \lt n\ $? I don't have any idea about that.
Can anybody please help me in this regard? Thanks in advance.
EDIT $:$ If $\phi$ was injective then $G$ can be thought of as a subgroup of $S_k.$ But since $n \gt \frac {k!} {2}$ it follows that $[S_k : G] =1,$ proving that $G = S_k.$ So $G$ has a non-trivial proper normal subgroup isomorphic to $A_k.$ In case $\phi$ is not injective then $\text {Ker}\ \phi$ gives a non-trivial proper normal subgroup of $G.$